- Spectral theorem
In

mathematics , particularlylinear algebra andfunctional analysis , the**spectral theorem**is any of a number of results aboutlinear operator s or about matrices. In broad terms the spectraltheorem provides conditions under which anoperator or a matrix can be diagonalized (that is, represented as adiagonal matrix in some basis). This concept of diagonalization is relatively straightforward for operators on finite-dimensional spaces, but requires some modification for operators on infinite-dimensional spaces. In general, the spectral theorem identifies a class oflinear operator s that can be modelled bymultiplication operator s, which are as simple as one can hope to find. In more abstract language, the spectral theorem is a statement about commutativeC*-algebra s. See alsospectral theory for a historical perspective.Examples of operators to which the spectral theorem applies are

self-adjoint operator s or more generallynormal operator s onHilbert space s.The spectral theorem also provides a canonical decomposition, called the

**spectral decomposition**,**eigenvalue decomposition**, or**eigendecomposition**, of the underlying vector space on which it acts.In this article we consider mainly the simplest kind of spectral theorem, that for a

self-adjoint operator on a Hilbert space. However, as noted above, the spectral theorem also holds for normal operators on a Hilbert space.**Finite-dimensional case****Hermitian matrices**We begin by considering a

Hermitian matrix "A" on a finite-dimensional real or complexinner product space "V" with the standard Hermitianinner product ; the Hermitian condition means:$langle\; A\; x\; ,,\; y\; angle\; =\; langle\; x\; ,,\; A\; y\; angle$

for all "x", "y" elements of "V".

An equivalent condition is that "A"* = "A", where "A"* is the

conjugate transpose of "A". If "A" is a real matrix, this is equivalent to "A"^{T}= "A" (that is, A is asymmetric matrix ). The eigenvalues of a Hermitian matrix are real.Recall that an

eigenvector of a linear operator "A" is a (non-zero) vector "x" such that "Ax" = λ"x" for some scalar λ. The value λ is the correspondingeigenvalue .**Theorem**. There is anorthonormal basis of "V" consisting of eigenvectors of "A". Each eigenvalue is real.We provide a sketch of a proof for the case where the underlying field of scalars is the

complex number s.By the

fundamental theorem of algebra , any square matrix with complex entries has at least one eigenvector. Now if "A" is Hermitian with eigenvector "e"_{1}, we can consider the space "K" = span{"e"_{1}}^{⊥}, the orthogonal complement of "e"_{1}. By Hermiticity, "K" is aninvariant subspace of "A". Applying the same argument to "K" shows that "A" has an eigenvector e_{2}∈ "K". Finite induction then finishes the proof.The spectral theorem holds also for symmetric matrices on finite-dimensional real inner product spaces, but the existence of an eigenvector is harder to establish. A real symmetric matrix has real eigenvalues, therefore eigenvectors with real entries.

If one chooses the eigenvectors of "A" as an orthonormal basis, the matrix representation of "A" in this basis is diagonal. Equivalently, "A" can be written as a linear combination of pairwise orthogonal projections, called its

**spectral decomposition**. Let:$V\_lambda\; =\; \{,v\; in\; V:\; A\; v\; =\; lambda\; v,\}$

be the eigenspace corresponding to an eigenvalue λ. Note that the definition does not depend on any choice of specific eigenvectors. "V" is the orthogonal direct sum of the spaces "V"

_{λ}where the index ranges over eigenvalues. Let "P"_{λ}be theorthogonal projection onto "V"_{λ}and λ_{1},..., λ_{"m"}the eigenvalues of "A", one can write its spectral decomposition thus::$A\; =lambda\_1\; P\_\{lambda\_1\}\; +cdots+lambda\_m\; P\_\{lambda\_m\}.$

The spectral decomposition is a special case of the

Schur decomposition . It is also a special case of thesingular value decomposition .**Normal matrices**The spectral theorem extends to a more general class of matrices. Let "A" be an operator on a finite-dimensional inner product space. "A" is said to be normal if "A"

^{*}"A" = "A A"^{*}. One can show that "A" is normal if and only if it is unitarily diagonalizable: By theSchur decomposition , we have "A" = "U T U"^{*}, where "U" is unitary and "T" upper-triangular. Since "A" is normal, "T T"^{*}= "T"^{*}"T". Therefore "T" must be diagonal. The converse is also obvious.In other words, "A" is normal if and only if there exists a

unitary matrix "U" such that:$A=U\; Lambda\; U^*\; ;$

where Λ is the

diagonal matrix the entries of which are theeigenvalue s of "A". The column vectors of "U" are the eigenvectors of "A" and they are orthonormal. Unlike the Hermitian case, the entries of Λ need not be real.**The spectral theorem for compact self-adjoint operators**In Hilbert spaces in general, the statement of the spectral theorem for compact

self-adjoint operators is virtually the same as in the finite-dimensional case.**Theorem**. Suppose "A" is a compact self-adjoint operator on a Hilbert space "V". There is anorthonormal basis of "V" consisting of eigenvectors of "A". Each eigenvalue is real.As for Hermitian matrices, the key point is to prove the existence of at least one nonzero eigenvector. To prove this, we cannot rely on determinants to show existence of eigenvalues, but instead one can use a maximization argument analogous to the variational characterization of eigenvalues. The above spectral theorem holds for real or complex Hilbert spaces.

If the compactness assumption is removed, it is not true that every self adjoint operator has eigenvectors.

**The spectral theorem for bounded self-adjoint operators**The next generalization we consider is that of bounded self-adjoint operators "A" on a Hilbert space "V". Such operators may have no eigenvalues: for instance let "A" be the operator multiplication by "t" on "L"

^{2}[0, 1] , that is:$[A\; varphi]\; (t)\; =\; t\; varphi(t).\; ;$

**Theorem**. Let "A" be a bounded self-adjoint operator on a Hilbert space "H". Then there is ameasure space ("X", Σ, μ) and a real-valued measurable function "f" on "X" and a unitary operator "U":"H" → "L"^{2}_{μ}("X") such that:$U^*\; T\; U\; =\; A\; ;$

where "T" is the

multiplication operator ::$[T\; varphi]\; (x)\; =\; f(x)\; varphi(x).\; ;$

This is the beginning of the vast research area of functional analysis called

operator theory .There is also an analogous spectral theorem for bounded normal operators on Hilbert spaces. The only difference in the conclusion is that now $f$ may be complex-valued.

An alternative formulation of the spectral theorem expresses the operator $A$ as an integral of the coordinate function over the operator's spectrum with respect to a

projection-valued measure . When the normal operator in question is compact, this version of the spectral theorem reduces to the finite-dimensional spectral theorem above, except that the operator is expressed as a linear combination of possibly infinitely many projections.**The spectral theorem for general self-adjoint operators**Many important linear operators which occur in analysis, such as

differential operators , are unbounded. There is also a spectral theorem forself-adjoint operator s that applies in these cases. To give an example, any constant coefficient differential operator is unitarily equivalent to a multiplication operator. Indeed the unitary operator that implements this equivalence is theFourier transform ; the multiplication operator is a type of Fourier multiplier.**ee also***

Matrix decomposition

*Canonical form

* Jordan decomposition, of which the spectral decomposition is a special case.

*Singular value decomposition , a generalisation of spectral theorem to arbitrary matrices.

*Eigendecomposition of a matrix **References*** Sheldon Axler, "Linear Algebra Done Right", Springer Verlag, 1997

*Paul Halmos , "What Does the Spectral Theorem Say?", "American Mathematical Monthly", volume 70, number 3 (1963), pages 241–247

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